电动力学每日一题 2021/10/12

(a) To make the EM field trapped inside a perfectly electric conducting cavity, the tangential component of the E-field must vanish on the internal surface:
E0J0(w0R/c)sin⁡(w0t)=0J0(w0R/c)=0E_0J_0(w_0 R/c)\sin(w_0 t)=0 \\ J_0(w_0R/c)=0E0J0(w0R/c)sin(w0t)=0J0(w0R/c)=0

This means the radius of cylinder must be chosen such that w0R/cw_0 R/cw0R/c is a zero of Bessel function J0(⋅)J_0(\cdot)J0(⋅).

(b) Maxwell 1st equation delievers the boundary condition of E-field
D⊥(r2,t)−D⊥(r1,t)=σS∓ϵ0E0J0(w0r∣∣/c)sin⁡(w0t)=σS(r∣∣,ϕ,±L/2,t)\textbf D_{\perp}(\textbf r_2,t)-\textbf D_{\perp}(\textbf r_1,t)=\sigma_S \\ \mp \epsilon_0 E_0 J_0(w_0 r_{||}/c)\sin(w_0t)=\sigma_S(r_{||},\phi,\pm L/2,t)D⊥(r2,t)−D⊥(r1,t)=σS∓ϵ0E0J0(w0r∣∣/c)sin(w0t)=σS(r∣∣,ϕ,±L/2,t)

(c) Maxwell 2nd equation delievers the boundary condition of H-field
H∣∣(r2,t)−H∣∣(r1)=Jfree(r0,t)×n^JS(R,ϕ,z,t)=−(E0/Z0)J1(w0R/c)cos⁡(w0t)z^JS(r∣∣,ϕ,±L/2,t)=±(E0/Z0)J1(w0r∣∣/c)cos⁡(w0t)z^\textbf H_{||}(\textbf r_2,t)-\textbf H_{||}(\textbf r_1) = \textbf J_{free}(\textbf r_0,t) \times \hat n \\ \textbf J_S(R,\phi,z,t)=-(E_0/Z_0)J_1(w_0R/c)\cos(w_0t)\hat z \\ \textbf J_S(r_{||},\phi,\pm L/2,t)=\pm (E_0/Z_0)J_1(w_0r_{||}/c)\cos(w_0 t)\hat zH∣∣(r2,t)−H∣∣(r1)=Jfree(r0,t)×n^JS(R,ϕ,z,t)=−(E0/Z0)J1(w0R/c)cos(w0t)z^JS(r∣∣,ϕ,±L/2,t)=±(E0/Z0)J1(w0r∣∣/c)cos(w0t)z^

(d) The charge-currrency continuinity equation is
∇⋅JS(r∣∣,ϕ,±L/2,t)+∂∂tσS(r∣∣,ϕ,±L/2,t)=0\nabla \cdot \textbf J_S(r_{||},\phi,\pm L/2,t) + \frac{\partial}{\partial t}\sigma_S(r_{||},\phi,\pm L/2,t)=0∇⋅JS(r∣∣,ϕ,±L/2,t)+∂t∂σS(r∣∣,ϕ,±L/2,t)=0

Evaluate
∇⋅JS=±(E0/Z0)∂r∣∣∂r∣∣r∣∣J1(w0r∣∣/c)cos⁡(w0t)=±(E0/Z0)J1(w0r∣∣/c)+(w0r∣∣/c)J1′(w0r∣∣/c)r∣∣cos⁡(w0t)=±(E0/Z0)(w0r∣∣/c)J0(w0r∣∣/c)r∣∣cos⁡(w0t)=±ϵ0E0w0J0(w0r∣∣/c)cos⁡(w0t)\begin{aligned}\nabla \cdot \textbf J_S & = \pm (E_0/Z_0) \frac{\partial }{r_{||} \partial r_{||}}r_{||}J_1(w_0 r_{||}/c)\cos(w_0 t) \\ & = \pm (E_0/Z_0) \frac{J_1(w_0 r_{||}/c)+(w_0 r_{||}/c)J_1'(w_0 r_{||}/c)}{r_{||}}\cos(w_0 t) \\ & = \pm (E_0/Z_0) \frac{(w_0 r_{||}/c)J_0(w_0 r_{||}/c)}{r_{||}}\cos(w_0 t) \\ & = \pm \epsilon_0 E_0 w_0 J_0(w_0 r_{||}/c)\cos(w_0 t)\end{aligned}∇⋅JS=±(E0/Z0)r∣∣∂r∣∣∂r∣∣J1(w0r∣∣/c)cos(w0t)=±(E0/Z0)r∣∣J1(w0r∣∣/c)+(w0r∣∣/c)J1′(w0r∣∣/c)cos(w0t)=±(E0/Z0)r∣∣(w0r∣∣/c)J0(w0r∣∣/c)cos(w0t)=±ϵ0E0w0J0(w0r∣∣/c)cos(w0t)

∂∂tσS=∂∂t∓ϵ0E0J0(w0r∣∣/c)sin⁡(w0t)=∓ϵ0E0J0(w0r∣∣/c)cos⁡(w0t)\begin{aligned} \frac{\partial}{\partial t}\sigma_S & = \frac{\partial}{\partial t} \mp \epsilon_0 E_0 J_0(w_0 r_{||}/c)\sin(w_0t) \\ & = \mp \epsilon_0 E_0 J_0(w_0 r_{||}/c)\cos(w_0t)\end{aligned}∂t∂σS=∂t∂∓ϵ0E0J0(w0r∣∣/c)sin(w0t)=∓ϵ0E0J0(w0r∣∣/c)cos(w0t)

Above,
∇⋅JS+∂∂tσS=0\nabla \cdot \textbf J_S+\frac{\partial}{\partial t}\sigma_S =0∇⋅JS+∂t∂σS=0

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